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I have an integral of this form:

$$v = \int_{x=-\infty}^\infty g(x) \delta \big(f(x, y_1) - f(x, y_2)\big) dx.$$

I know that if $y_1 \neq y_2$, a closed form solution can be found, namely $$ v = \frac{g(\hat{x})}{\left|\frac{\partial f }{\partial x}(\hat{x}, y_1) - \frac{\partial f }{\partial x}(\hat{x}, y_2) \right|},$$ with $\hat{x}$ the solution of $f(x, y_1) - f(x, y_2)=0$ (supposed unique).

My question may seem weird, but I am wondering what happen when $y_1=y_2$ ? The integral diverges, I know, but can we actually still say something on its solution? For example, is there a way to "normalize" the solution by infinity ? I don't know if it even makes sense and, if it does, I don't know how to formalize this idea. Any help would be appreciated, thank you.

Edit: thank you for the comments/answers, it cleared things up. I had a specific case where the first equation is actually used to define a linear operator, and I realize now that the question cannot be separated from this context. After further investigations, things actually check out without an infinity term appearing.

Denys
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  • If $y_1=y_2$, then $f(x,y_1)-f(x,y_2)\equiv 0$, and the problem is ill-posed. In fact, even if $y_1\ne y_2$, but $f(x,y_1)-f(x,y_2)\equiv 0$, the problem is ill-posed, – Mark Viola Sep 21 '22 at 16:53
  • $f(x,y_1)-f(x,y_2)$ here is just a function of $x$ (albeit parameterized by $y_1$ and $y_2$); call it $h(x)$. If it has isolated zeroes at $x_i$, and all $h'(x_i)\neq 0$, then $\int g(x)\delta(h(x))dx = \sum_i g(x_i) / |h'(x_i)|$. Otherwise the integral diverges. There may be an interesting edge case when $g(x_i)=0$ and, say, $h(x_i + \epsilon) \sim \epsilon^2$. – mjqxxxx Sep 21 '22 at 17:23

1 Answers1

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You know that

$$\int_{-\infty}^{+\infty} g(x)\delta(x)\ \text{d}x = g(0)$$

In the sense of distributions. But if you put $\delta(0)$ into the integrand, then this stop making sense for

$$\int_{-\infty}^{+\infty} g(x)\delta(0) \ \text{d}x = \delta(0)\int_{-\infty}^{+\infty} g(x)\ \text{d}x$$

The integral could diverge, and besides this, $\delta(0) = ??$

Defining $\delta(0)=\infty$ is not at all right, because it has to equal $\infty$ in just the "right way". You can define this as a distribution, which ignoring all technicalities, just means that it is an "evaluation linear map". I.e, given an appropriate vector space $V$ of functions $\Bbb{R}^n\to\Bbb{C}$, we consider $\delta_a:V\to\Bbb{C}$ defined as $\delta_a(f):= f(a)$. In other words, $\delta_a$ is an object which takes a function as input and gives out the value of the function a the point $a$.

When we write $$\int_{\Bbb{R}^n}\delta_a(x)f(x)\,dx=\int_{\Bbb{R}^n}\delta(x-a)f(x)\,dx=f(a)$$

we literally mean that $\delta_a$ is an object such that whenever we plug in a function $f$, we get out its value at $a$. Of course, writing it in this way inside an integral is a priori just nonsense, because there is no function $\delta_a:\Bbb{R}^n\to\Bbb{C}$ for which the above equality can hold true.

Trying to give a sense to $\delta(0)$ is neither easy nor meaningful, for what I know. The same problem appears in realtivistic quantum mechanics when, before passing to the second quantisation, you obtain that the total energy of a quantum field is a sum of dirac delta in zero, that is ill-posed and ill-defined.

I am answering by heart so I will perhaps add some details later.

I also leave you this, for possible interest: https://core.ac.uk/download/pdf/82055462.pdf

Enrico M.
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    (+1) for trying to make meaning out of something meaningless. It is amazing that the Dirac Delta is still being taught incorrectly. – Mark Viola Sep 21 '22 at 18:00
  • @MarkViola Ehm, didn't really understand if that comment aimed to say that my answer is wrong and poor or if it was a general comment... >.< – Enrico M. Sep 21 '22 at 19:27
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    The response is neither wrong nor poor. It's just that the question is ... well, meaningless. – Mark Viola Sep 21 '22 at 19:53
  • @MarkViola Oh, goti it! I do agree. Yet I always try to put myself in those people's shoes. Most of them do not know (yet?) the majority of the mathematics they will hopefully learn in the next years, so they might jsut want to know! Sometimes they hurt hitting a wall, other times they break one :D – Enrico M. Sep 21 '22 at 20:04
  • Oops ... Forgot to actually give the +1 – Mark Viola Sep 21 '22 at 23:40
  • "It is amazing that the Dirac Delta is still being taught incorrectly". Indeed, this is quite sad ... I again put a big answer here https://math.stackexchange.com/questions/4538645/prove-properties-of-dirac-delta-from-the-definition-as-a-distribution that might complete your answer ;) – LL 3.14 Sep 26 '22 at 20:23